# Solution conflict!

by gulsen
Tags: conflict, solution
 P: 218 I have a basic differential equation: $$\frac{dy}{dx} = x + y, y(0) = 1$$ Now, when I try to solve this by making it exact $$\mu \frac{dy}{dx} + \mu y = \mu x$$ I get $$\mu = e^{-x}$$ and solution $$-x-1$$. This doesn't satisfy the initial condition. But when I try to solve it as a non-homogenous equation as: $$\frac{dy}{dx} + y= x$$ I get $$y_p = 2e^x, y_c = -x-1$$ so $$y = 2e^x-x-1$$ Which seems to be a correct & full solution. What was I missing in the first try?
 P: 74 You are solving dy/dx + y = x which is wrong since the original equation is dy/dx = x + y, or dy/dx - y = x so now u(x) = e^(-x) and so on and it works out. And yes, the answer is y = 2e^x - x - 1
Mentor
P: 15,202
 Quote by gulsen I have a basic differential equation: $$\frac{dy}{dx} = x + y, y(0) = 1$$ Now, when I try to solve this by making it exact $$\mu \frac{dy}{dx} + \mu y = \mu x$$
I assume this was just a typo, but that equation should be
$$\mu \frac{dy}{dx} - \mu y = \mu x$$
 I get $$\mu = e^{-x}$$ and solution $$-x-1$$. This doesn't satisfy the initial condition.
Your error: you forgot the constant of integration. You should have obtained
$$y = -(x+1) + c/\mu = ce^x - (1+x)$$
and then solved for the initial condition $y(0)=1$ yielding $c=2$ or
$$y = 2e^x -(x+1)$$
 But when I try to solve it as a non-homogenous equation as: $$\frac{dy}{dx} + y= x$$ I get $$y_p = 2e^x, y_c = -x-1$$ so $$y = 2e^x-x-1$$
You're nomenclature is backward here. The solution to the homogeneous equation is called the complementary function and is denoted as $y_c$. The complementary function generally involves arbitrary constants. A solution to the inhomogeneous equation is called a particular function and is denoted as $y_p$.
In this case, the solution to the homogeneous equation $y^\prime-y=0$ is
$$y_c = ce^x$$
where $c$ is an arbitrary constant and
$$y_p = -(x+1)$$
is a particular solution to the inhomogeneous equation. Combining these,
$$y = ce^x - (1+x)$$
which meets the initial conditions when $c=2$.

 P: 218 Solution conflict! Oh, that was just a typo. Thanks D_H, it ends up that I've forgotten the integration constant. About naming, take it easy, I'm just a freshman!

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